## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can divide the motion into three distances $x_1$, $x_2$, and $x_3$; $x_1 = \frac{1}{2}at^2$ $x_1 = \frac{1}{2}(4.0~m/s^2)(6.0~s)^2$ $x_1 = 72~m$ After 6.0 seconds, the speed is $(4.0~m/s^2)(6.0~s)$ which is 24 m/s. Therefore; $x_2 = v~t = (24~m/s)(2.0~s) = 48~m$ $x_3 = \frac{v^2-v_0^2}{2a}$ $x_3 = \frac{0-(24~m/s)^2}{(2)(-3.0~m/s^2)}$ $x_3 = 96~m$ We can then find the total distance $x$ between the stop signs: $x = x_1+x_2+x_3$ $x = 72~m+48~m+96~m$ $x = 216~m$ The distance between the stop signs is 216 meters.